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Len [333]
3 years ago
12

Higher Order Thinking A yellow ribbon

Mathematics
1 answer:
Olegator [25]3 years ago
8 0

Answer:

112

Step-by-step explanation:

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Write two fractions that have the same value as 1/2.Explain how you know they have the same value
BaLLatris [955]

Answer:

2/4 3/6  and more

Step-by-step explanation:

the reason is that for and example 2/4 if you divide both the numerator (top) and the denominator (bottom) by 2 it equals 1/2. So 2 divided by 2 is 1 and 4 divided by 2 is 2 . Another example is 3/6 because 3 divided by 3 is 1 and 6 divided 3 is 2.

7 0
3 years ago
How do you solve? using either factoring, completing the square, or quadratic formula?
lina2011 [118]
X=sq root of 11; x=-sq root of 11
6 0
2 years ago
Read 2 more answers
Solve for n.<br> n + 1 = 4(n – 8)<br> n=1<br> n=8<br> n=11<br> U n = 16
Alika [10]

\displaystylen+1=4(n-8) \\n+1=4n-32 \\-3n+33=0 \\-3(n-11)=0 \\n-11=0 \\n=11

Hope this helps.

3 0
3 years ago
Read 2 more answers
Joseph has $60 in a savings account. Each week he deposits $12 in the account.
Aleonysh [2.5K]

Answer:

12x +60

Step-by-step explanation:

c

d

5 0
2 years ago
You decide to put $5000 in a savings account to save $6000 down payment on a new car. If the account has an interest rate of 7%
bezimeni [28]

\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill&\$6000\\ P=\textit{original amount deposited}\dotfill &\$5000\\ r=rate\to 7\%\to \frac{7}{100}\dotfill &0.07\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12\\ t=years \end{cases}

\bf 6000=5000\left(1+\frac{0.07}{12}\right)^{12\cdot t}\implies \cfrac{6000}{5000}\approx (1.0058)^{12t}\implies \cfrac{6}{5}\approx(1.0058)^{12t} \\\\\\ \log\left( \cfrac{6}{5} \right)\approx \log[(1.0058)^{12t}]\implies \log\left( \cfrac{6}{5} \right)\approx 12t\log(1.0058) \\\\\\ \cfrac{\log\left( \frac{6}{5} \right)}{12\log(1.0058)}\approx t\implies 2.63\approx t\impliedby \textit{about 2 years, 7 months and 16 days}

6 0
3 years ago
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